Question:
If $z_{1}=(1+i)$ and $z_{2}=(-2+4 i)$, prove that $\operatorname{lm}\left(\frac{z_{1} z_{2}}{z_{1}}\right)=2$
Solution:
We have, $z_{1}=(1+i)$ and $z_{2}=(-2+4 i)$
Now, $\frac{z_{1} z_{2}}{\overline{z_{1}}}=\frac{(1+\mathrm{i})(-2+4 \mathrm{i})}{(1+1)}$
$=\frac{-2+4 i-2 i+4 i^{2}}{(1-i)}=\frac{-2+4 i-2 i-4}{(1-i)}=\frac{-6+2 i}{(1-i)}$
$=\frac{-6+2 i}{(1-i)} \times \frac{(1+i)}{(1+i)}$
$=\frac{-6-6 i+2 i+2 i^{2}}{1+1}$
$=\frac{-6-4 i-2}{2}=\frac{-8-4 i}{2}$
$=-4-2 i$
Hence, $\operatorname{Im}\left(\frac{z_{1} z_{2}}{z_{2}}\right)=-2$